Exercise :

Find the group of symmetries of the Cube.

Attempt :

The elements are:

$3$ rotations (by $\pi/2$ or $\pi$) about the centers of $3$ pairs of opposite faces.

$1$ rotation (by $\pi$) about the centers of $6$ pairs of opposite edges.

$2$ rotations (by $2\pi/3$) about $4$ pairs of opposite vertices (diagonals).

Together with the identity this accounts for all $24$ elements (the order of the group of direct symmetries is $24$).

Every rotation determines a permutation of the four diagonals and this defines the isomorphism, which tells us that the group of symmetries of the Cube is the group $S_4$.

Question :

Is my approach correct? This is an exercise I have to solve as part of an academic list of problems, so I want to make sure it's perfect and strict mathematically. If not, please tell me with details how it should be.

  • $\begingroup$ I think you're missing the isometries of the cube which do not correspond to physical motions in three space. Isometries also include those transformations which amount to turning the cube inside out. Taking those into account there are 48 symmetries. So, what's your definition of "symmetry"? $\endgroup$ – James S. Cook Nov 7 '17 at 20:50
  • $\begingroup$ For rotational symmetries, it is $S_4$. One thing to keep in mind is how to argue that your list of symmetries is exhaustive. P.S. you can geometrically see that $A_4 \subset S_4$ from the fact that a tetrahedron can be inscribed within a cube. $\endgroup$ – Kaj Hansen Nov 7 '17 at 20:53
  • $\begingroup$ @KajHansen How can I prove that these symmetries are unique ? Is my solution missing anything except from that ? I'd really appreciate if you could show me how a complete solution would be. $\endgroup$ – Rebellos Nov 7 '17 at 20:54
  • $\begingroup$ @JamesS.Cook Didn't really grasp why it became 48 ! I'm a beginner on this course, so don't mind me ! $\endgroup$ – Rebellos Nov 7 '17 at 20:54
  • $\begingroup$ I have to vote here shortly, so I'll be away a while. I discuss "symmetry" verses "motion" (aka a motion is a rotational symmetry) around minute 10 of youtube.com/… I'm sure Kah Hansen will set you straight in my absence. $\endgroup$ – James S. Cook Nov 7 '17 at 20:56

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